The difference is between $\mbox{sinc}$ and the periodic version obtained using the DFT.

See this answer for a comparison.

It strikes me that asinc = ratio of two sincs.

The $\mbox{asinc}$ function is a ratio of two $\sin$ functions: $$ \frac{\sin(N\omega/2)}{\sin(\omega/2)} = \frac{\sin(N\omega/2)}{\omega/2} \frac{\omega/2}{\sin(\omega/2)} $$ so, yes, you could see it as the ratio of two $\mbox{sinc}$ functions.

Underlying question: can I model the sampling of a continuous signal (by, say, a digital camera) in the frequency domain as the DFT of the signal convolved with a Dirac delta train times the DFT of the pixel footprint (a rect)? If so, why is the DFT of the pixel footprint an asinc instead of a sinc?

So you seem to be trying to model taking a picture with a digital camera? Camera modeling is a little tricky. Even Gonzales and Woods seem to gloss over it.

The fundamental issue is the DFT of a $\mbox{rect}$ ($\Pi$) is a $\mbox{asinc}$.

If you're doing a discrete-time Fourier transform (DTFT), then it's not, but usually when dealing with computed FTs, you want the DFT.

The difference is between $\mbox{sinc}$ and the periodic version obtained using the DFT.

See this answer for a comparison.

It strikes me that asinc = ratio of two sincs.

The $\mbox{asinc}$ function is a ratio of two $\sin$ functions: $$ \frac{\sin(N\omega/2)}{\sin(\omega/2)} = \frac{\sin(N\omega/2)}{\omega/2} \frac{\omega/2}{\sin(\omega/2)} $$ so, yes, you could see it as the ratio of two $\mbox{sinc}$ functions.

Underlying question: can I model the sampling of a continuous signal (by, say, a digital camera) in the frequency domain as the DFT of the signal convolved with a Dirac delta train times the DFT of the pixel footprint (a rect)? If so, why is the DFT of the pixel footprint an asinc instead of a sinc?

So you seem to be trying to model taking a picture with a digital camera? Camera modeling is a little tricky. Even Gonzales and Woods seem to gloss over it.

The fundamental issue is the DFT of a $\mbox{rect}$ ($\Pi$) is a $\mbox{asinc}$.

If you're doing a discrete-time Fourier transform (DTFT), then it's not, but usually when dealing with computed FTs, you want the DFT.

The difference is between $\mbox{sinc}$ and the periodic version obtained using the DFT.

See this answer for a comparison.

The difference is between $\mbox{sinc}$ and the periodic version obtained using the DFT.

See this answer for a comparison.

It strikes me that asinc = ratio of two sincs.

The $\mbox{asinc}$ function is a ratio of two $\sin$ functions: $$ \frac{\sin(N\omega/2)}{\sin(\omega/2)} = \frac{\sin(N\omega/2)}{\omega/2} \frac{\omega/2}{\sin(\omega/2)} $$ so, yes, you could see it as the ratio of two $\mbox{sinc}$ functions.

Underlying question: can I model the sampling of a continuous signal (by, say, a digital camera) in the frequency domain as the DFT of the signal convolved with a Dirac delta train times the DFT of the pixel footprint (a rect)? If so, why is the DFT of the pixel footprint an asinc instead of a sinc?

So you seem to be trying to model taking a picture with a digital camera? Camera modeling is a little tricky. Even Gonzales and Woods seem to gloss over it.

The fundamental issue is the DFT of a $\mbox{rect}$ ($\Pi$) is a $\mbox{asinc}$.

If you're doing a discrete-time Fourier transform (DTFT), then it's not, but usually when dealing with computed FTs, you want the DFT.